Domain of the infinitesimal generator of the shift semigroup

Question

In a course at my university, we study strongly continuous semigroups and their infinitesimal generators. In a simple example, we take a look at a shift semigroup. let $ T $ be an operator on $ X = L^2(0, 1) $, defined by:

if $ t + x \leq 1 $, $ (T(t)f)(x) = f(t + x) $
if $ t + x > 1 $, $ (T(t)f)(x) = 0 $

for $t \geq 0$

So $ T(t) $ basically takes a square integrable function on (0, 1) and shifts it to the left $ t $ units, filling the space that shifts in from the right with 0.

When $ \lim_{t \downarrow 0} \frac{T(t)f - f}{t} $ exists in $ X $, we say that $f \in D(A) $ and we define the infinitesimal generator $ A $ by setting $ Af $ to the limit above.

When I try to find the domain of the infinitesimal generator $A$, $ D(A) $, I come to the conclusion that we have $ A = \frac{\text{d}}{\text{dx}}$, and we must have f(1) = 0 in order for the limit to exist when $ x + t = 1 $. Furthermore, we need $ f $ to be continuous and differentiable (in order for the limit to exist), and $ f' $ to be in $ X = L^2(0, 1) $ (in order for the derivative to be in $ X $).

This seems fine to me, but in the lecture and other examples I found online, it is claimed that the functions in $ D(A) $ need to be absolutely continuous as well. I am not really familiar with the concept of absolute continuity, and I can't find an explanation or reason why we need this extra condition on the functions in $ D(A) $ (this part is suspiciously left as an exercise for the reader, or skipped in a sneaky way in all sources I've found).

Answer 1

Absolute continuity is a more general concept than differentiability, i.e. not every absolutely continuous function is differentiable. However, the following statements hold true:

• Any continuously differentiable function on, say $[a,b]$, is absolutely continuous.
• Any absolutely continuous function is almost everywhere differentiable.

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Suppose that $f \in L^2(0,1)$ is absolutely continuous. Then it is almost everyhwere differentiable and therefore

$$g(x) := \lim_{t \to 0} \frac{T_t f(x)-f(x)}{t}$$

exists for almost all $x$. Since $L^2$-functions are only defined up to exceptional null sets, we can set $g(x) := 0$ for all $x$ such that the above limit does not exist. Then

$$\lim_{t \to 0} \frac{T_t f-f}{t} = g \in L^2 \qquad \text{almost everywhere}.$$